TypeScript Version: 2.4.1

Code

// strictNullChecks: true let x : string | undefined let y = x! type t1 = typeof x! type t2 = typeof y

Expected behavior:
the t1 statement should create a type of type string.
the t2 statement should create a type of type string.

Actual behavior:
the t1 statement errors with ';' semicolon expected.
the t2 statement creates a type of type string.

There appears to be no way to achieve this without defining a separate variable, or by type guarding to remove the undefined / null.

With strict null checks on, this is especially troublesome in the case of generics with extends clauses, i.e.

class Foo {} class Bar<T extends Foo> {} const x : Foo | undefined const y = new Bar<typeof x>() // throws an exception because typeof x === Foo | undefined != Foo

It would be great if the non-null assertion operator worked natively in typeof statements, or even if you could use brackets to resolve the assertion before the typeof.